Evaluate the improper integral if it exists. $\int^{\infty}_{2}\dfrac{1}{\sqrt{x}}\,dx $ Choose 1 answer: Choose 1 answer: (Choice A) A $0$ (Choice B) B $2$ (Choice C) C $2\sqrt{2}$ (Choice D) D The improper integral diverges.
Explanation: First, let's rewrite the improper integral: $\int_{2}^{\infty}\dfrac1{\sqrt x}\,dx=\lim_{b\to\infty}\int_{2}^{b}\dfrac1{\sqrt x}\,dx$ We can now evaluate the integral: $\begin{aligned} \phantom{\int_{2}^{\infty}\dfrac1{\sqrt x}\,dx}&=\lim_{b\to\infty}\int_{2}^{b}\dfrac1{\sqrt x}\,dx\\ \\ \\ &=\lim_{b\to\infty}\Big[2\sqrt x\Big]_{2}^b\\ \\ \\ &=\lim_{b\to\infty}\left(2\sqrt b-2\sqrt2\right)\\ \\ &=\lim_{b\to\infty}2\sqrt b-\lim_{b\to\infty}2\sqrt{2}\\ \\ &=\infty-2\sqrt{2}\\ \\ &=\infty \end{aligned}$ The answer: The improper integral diverges.